A circular disc of moment of inertia `I_(t)` is rotating in a horizontal plane about its symmetry axis with a constant angular velocity `omega_(i)`. Another disc of moment of inertia `I_(b)` is dropped co-axially onto the rotating disc. Initially, the second disc has zero angular speed. Eventually, both the discs rotate with a constant angular speed `omega_(f)`. Calculate the energy lost by the initially rotating disc due to friction.

Options

A. `1/2(I_(b)^(2))/(I_(t)+I_(b))omega_(t)^(2)`
B. `1/2 (I_(t)^(2))/((I_(t) +I_(b))) omega_(i)^(2)`
C. `1/2 (I_(b)-I_(t))/((I_(t)=I_(b))) omega_(i)^(2)`
D. `1/2 (I_(b)I_(t))/((I_(t)+I_(b)))omega_(i)^(2)`

Answer

D

Explanation

Loss of energy
` Delta E = 1/2 I_(t) omega_(t)^(2) - 1/2 (I_(t)omega_(i)^(2))/((I_(t)+I_(b))) = 1/2 (I_(b) I_(t)omega_(i)^(2))/((I_(t)+I_(b)))`